Jabarkan bentuk aljabar dari (2x + 3)3 dan (3x – 5y)4
Jawaban
kita dapat gunakan pola Segitiga Pascal:
Berikut pola bilangan Segitiga Pascal:
1 → (a + b)⁰ = 1
1 1 → (a + b)¹ = a + b
1 2 1 → (a + b)² = a + 2ab + b²
1 3 3 1 → (a + b)³ = a³ + 3a²b + 3ab² + b³
1 4 6 4 1 → (a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴
1 5 10 10 5 1 → (a + b)⁵ = a⁵ + 5a⁴b + 10a³b² + 10a²b³ + 5ab⁴ + b⁵
=> (2x + 3)³ gunakan pola → (a + b)³ = a³ + 3a²b + 3ab² + b³
a = 2x dan b = 3, diperoleh:
(2x + 3)³ = (2x)³ + 3(2x)²(3) + 3(2x)(3)² + 3³
= 2³x³ + 9(2²x²) + (6x)(9) + 27
= 8x³ + 9(4x²) + 54x + 27
= 8x³ + 36x² + 54x + 27
=> (3x – 5y)⁴ gunakan pola → (a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴
a = 3x dan b = -5y
(3x – 5y)⁴ = (3x)⁴ + 4(3x)³(-5y) + 6(3x)²(-5y)² + 4(3x)(-5y)³ + (-5y)⁴
= 3⁴x⁴ – 20y(3³x³) + 6(3²x²)((-5)²y²) + 12x((-5)³y³) + 5⁴y⁴
= 81x⁴ – 20y(27x³) + 6(9x²)(25y²) + 12x(-125y³) + 625y⁴
= 81x⁴ – 540x³y + 1350x²y² – 1500xy³ + 625y⁴
